Question

Find the Limit:

Answer 1

\[\lim_{x \rightarrow \infty}(\frac{ 5x-3 }{ 5x+7 })^{3x+1}\]

Answer 2

take the log
take the limit
exponentiate

Answer 3

I am allowed to use L'Hospital's rule. I am not sure how I would use it though...

Answer 4

So I get it's (infinity/infinity)^(infinity) but I am not sure how to go from there...

Answer 5

Yep... I am not sure how that would help though...

Answer 6

\[(3x+1)\ln(\frac{5x-3}{5x+7})\]
now you \(\infty\times 0\)

Answer 7

Yep... I got there.

Answer 8

rewrite as
\[\frac{\ln(\frac{5x-3}{5x+7})}{\frac{1}{3x+1}}\] now you have
\[\frac{0}{0}\] and now you can use l'hopital

Answer 9

Hmm Okay... Let's see...

Answer 10

probably easier if you write
\[\frac{\ln(5x-3)-\ln(5x+7)}{\frac{1}{3x+1}}\]

Answer 11

Okay Thanks!

Answer 12

Yep. I know.

Answer 13

yw

Answer 14

@satellite73 : Okay so I have:

\[\lim_{x \rightarrow \infty} \frac{ \frac{ 5 }{ 5x-3 }-\frac{ 5 }{ 5x+7 } }{ \frac{ -2 }{ (3x+1)^2 } }\]

Answer 15

denominator is wrong

Answer 16

\[\lim_{x \rightarrow \infty} \frac{ \frac{ 5 }{ 5x-3 }-\frac{ 5 }{ 5x+7 } }{ \frac{ -3 }{ (3x+1)^2 } }\]

Answer 17

Chain rule. Right.

Answer 18

ugly retricealgebra from here on in
have fun, but you will get a rational expression so limit is same as horizontal asymptote

Answer 19

:( . Fair enough....

Answer 20

actually it is not that bad

Answer 21

Yep. I think I see why.

Answer 22

\[-\frac{(3x+1)^2}{3}\times \frac{50}{(5x-3)(5x+7)}\] and not need to multiply out, just look at the ratio of the leading coefficients

Answer 23

i mean you have to be a bit careful but the leading coefficient of the numerator is \(9\times 50\) and the leading coefficient of the denominator is \(-3\times 25\) so you the ratio is \(-6\)

Answer 24

and don't forget to exponentiate at the end

Answer 25

Hmm... Okay.

Answer 26

So I got 1/e^6 or e^-6.

Answer 27

yup

Answer 28

:) . Thanks a million :) .